Let's solve the equation 2k^2 = 9 + 3k
First, subtract each side by (9+3k) to get 0 on the right side of the equation
2k^2 = 9 + 3k
2k^2 - (9+3k) = 9+3k - (9+3k)
2k^2 - 9 - 3k = 9 + 3k - 9 - 3k
2k^2 - 3k - 9 = 0
As you see, we got a quadratic equation of general form ax^2 + bx + c, in which a = 2, b= -3, and c = -9.
Δ = b^2 - 4ac
Δ = (-3)^2 - 4 (2)(-9)
Δ<u /> = 9 + 72
Δ<u /> = 81
Δ<u />>0 so the equation got 2 real solutions:
k = (-b + √Δ)/2a = (-(-3) + √<u />81) / 2*2 = (3+9)/4 = 12/4 = 3
AND
k = (-b -√Δ)/2a = (-(-3) - √<u />81)/2*2 = (3-9)/4 = -6/4 = -3/2
So the solutions to 2k^2 = 9+3k are k=3 and k=-3/2
A rational number is either an integer number, or a decimal number that got a definitive number of digits after the decimal point.
3 is an integer number, so it's rational.
-3/2 = -1.5, and -1.5 got a definitive number of digit after the decimal point, so it's rational.
So 2k^2 = 9 + 3k have two rational solutions (Option B).
Hope this Helps! :)
Point-slope form: y-y1 = m(x-x1)
Standard form: ax + by = c
Slope-intercept form: y = mx+b
Start by finding the slope. We know it is negative since the line is decreasing. The slope is -4/3.
To create point-slope form, we need to get one point from the graph. Let's use (3,0).
To create slope-intercept form, we need the slope and the y-intercept. The y-intercept is the point where our equation crosses the y-axis. For this equation, it is 4.
To get standard form, solve the equation in terms of C.
Point-slope form: y = -4/3(x-3)
Slope-intercept form: y = -4/3x + 4
Standard form: 4/3x + y = 4
Thank you very much and this “question” was very helpful for me.
